**Section 1.6 **

**Homogeneous Systems, Subspaces, and Bases**

`> `

`> `

**Homogeneous Systems**

A
__homogeneous system__
of m equations with n unknowns x
*i*
(
*i = 1.. n*
) is a system of

the form

+ ........ + = 0

+ ........ + = 0

+ ........ + = 0

.............

+ ........ + = 0

Is a homogeneous system consistent or inconsistent?

**Answer**

The homogeneous system of linear equations is always consistent.

The solution set of a homogeneous system is never empty. In fact,
, ..
always satisfy such systems. This solution is called the
**zero **
or
**trivial solution**
.

******************************************************************

**Example 1:**
Consider the system:

`> `
**eq1:=x[1]+x[2]+2*x[3]=0: eq2:=2*x[1]+x[2]-3*x[3]=0: print(eq1); print(eq2);**

`> `

What is the solution set of this system?

`> `
**solveqns({eq1,eq2},{x[1],x[2],x[3]});**

Select one of the following modes

1. Demonstration mode

2. No-Step mode

`
`**2;**

`> `

`> `

`> `

`> `

`> `

`> `

`> `

******************************************************************

Example 1 represents a system with infinitely many solutions and with one free variable. ******************************************************************

**Example 2**
Consider the system:

`> `
**eq1:=x[1]+2*x[2]-5*x[3]+6*x[4]=0: print(eq1);**

`> `

What is the solution set of this system?

`> `
**solveqns(eq1,{x[1],x[2],x[3],x[4]});**

Select one of the following modes

1. Demonstration mode

2. No-Step mode

`
`**2;**

`> `

`> `

********************************************************************

This system has infinitely many solutions with three free variable.

********************************************************************

**Example 3:**
Consider the system:

`> `
**eq1:=x[1]-2*x[2]-x[3]=0: eq2:=2*x[1]-x[2]+x[3]=0: eq3:=x[1]+3*x[2]-5*x[3]=0:**

`> `
**print(eq1); print(eq2); print(eq3);**

What is the solution set of this system?

`> `
**solveqns({eq1,eq2,eq3},{x[1],x[2],x[3]});**

Select one of the following modes

1. Demonstration mode

2. No-Step mode

`
`**2;**

`> `

`> `

`> `

`> `

`> `

`> `

`> `

******************************************************************

This system has the trivial solution as its unique solution.

******************************************************************

Examples illustrate several scenarios for homogeneous systems.

*******************************************************************

If the number of equations is fewer than the number of unknowns (
__under determined system__
),

then the homogeneous system will have infinitely many solutions.

If the number of equations is greater or equal to the number of unknowns (
__over determined__

__system__
), then the homogeneous system will have either a unique solution or infinitely many solutions.

A homogeneous systems of linear equations is consistent. The solution set is either a

singleton or infinite.

***********************************************************************

Every linear combination of solutions of a homogeneous system Ax=0 is again a solution of the system.

**Nonhomogeneous Systems**

Consider the
__nonhomogeneous__
system of m equations with n unknowns:

+ ........ + =

+ ........ + =

+ ........ + =

.............

+ ........ + =

*************************************************************************

Every system of linear equations Ax = b has either no solution, exactly one solution or

infinitely many solutions.

**Example 4:**
Consider the system:

`> `
**eq1:=x[1]+x[2]+2*x[3]=1: eq2:=2*x[1]+2*x[2]-4*x[3]=2:**

`> `
**print(eq1); print(eq2);**

`> `

What is the solution set of this system?

`> `
**solveqns({eq1,eq2},{x[1],x[2],x[3]});**

Select one of the following modes

1. Demonstration mode

2. No-Step mode

`
`**2;**

`> `

`> `

`> `

`> `

`> `

`> `

`> `

`> `

********************************************************************

This system has infinitely many solutions with one free variable. The solution of the

nonhomogeneous system can be expressed as

`> `
**matrix(3,1,[x[1],x[2],x[3]]) = matrix(3,1,[1,0,0]) + a*matrix(3,1,[-1,1,0]);**

where
is the free variable.The solution [1,0,0] is called the
** **
__particular solution__** **
to the system; while
is a solution of the associated homogeneous system.

*********************************************************************

**Example 5:**
Consider the system:

`> `
**eq1:=x[1]+2*x[2]-5*x[3]+6*x[4]=10: print(eq1);**

`> `

What is the solution set of this system?

`> `
**solveqns(eq1,{x[1],x[2],x[3],x[4]});**

Select one of the following modes

1. Demonstration mode

2. No-Step mode

`
`**2;**

`> `

`> `

`> `

This system has infinitely many solutions with
__three free variables__
. The general solution

[ ] of the nonhomogeneous system is

=

=

The solution [10,0,0,0] is the particular solution of the system; while the solutions

; ; and are solutions to the associated

homogeneous system.

***********************************************************************

***********************************************************************

Let
*Ax=b*
be a linear system. If
*p*
is any particular solution of
*Ax=b *
and
*h*
is a solution of the

corresponding homogeneous system
*Ax=0*
, then
*p+h*
is a solution of
*Ax=b*
. Moreover, every

solution of
*Ax=b*
has tghe form
*p+h,*
so that the general solution is
*x=p+h*
where
*Ah=0*
.

***********************************************************************

**Subspaces**

*********************************************************************

A subset
*W *
of
is closed under vector addition if for all
*u,v*
in
*W*
the sum
* u+v*
is in
*W*
.

If r
*v*
is in
*W *
for all
*v*
in
*W*
and all scalars r, then
*W*
is closed under scalar multiplication.

A non-empty subset
*W*
of
that is closed under both vector addition and scalar

multiplication is a
__subspace__
of
.

*********************************************************************

**Example 6:**

`> `
**`S1 ={[x,y,z] | 2*x+3*y+z=10}`;**

Choose two vectors in :

`> `
**v:=vector([-4,3,9]);u:=vector([-7,4,12]);**

Compute their sum

`> `
**`v+u`=evalm(v+u);**

Does vector v + u belong to subset ? Why?

**Example 7:**

`> `
**`S2 ={[x,y,z] | 2*x+3*y+z=0}`;**

`> `

Is subset closed under addition? under scalar multiplication?

**Answer**

Subset
is closed under the two operations. Hence
is a
**subspace**
of
.

`> `
**with(plots):**

Warning, the name changecoords has been redefined

`> `
**implicitplot3d({2*x+3*y+z=0,2*x+3*y+z=10},x=-5..5,y=-5..5,z=0..10,axes=normal);**

`> `

We observe that the graph of passes through the origin while the graph of does not.

The zero vector belongs to the set but does not belong to the set .

Graphically, can you tell which subset does not contain the zero vector?

What is the implication of not having the zero vector in the set?

**Example 8: **
Consider the following set

`> `
**`S1 = {[x,y] | x^2 + y^2 <= 1}`;**

Note that the set {0} and the set are subspaces of .

*****************************************************************

Let W = sp{ , . . ., } be the span of k vectors in . Then W is a subspace

of .

******************************************************************

**Example 9:**
Consider the homogeneous system of linear equations

`> `
**eq1:=x1 + 3*x2 + 5*x3 = 0:
eq2:=3*x1+ 2*x2 + 5*x3 = 0:**

`> `
**print(eq1); print(eq2);**

The system is defined by matrix multiplication A*x = 0 where A and x are

`> `
**A:=matrix([[1,3,5],[3,2,5]]);
x:=matrix([[x1],[x2],[x3]]);**

`> `

which is equivalent to

`> `
**multiply(A,x)=0;**

`> `

Upon solving the system

`> `
**solve({eq1,eq2},{x1,x2,x3});**

`> `

we obtain

`> `
**`S ={[x1,x2,x3]|x1=(-5/7)*a,x2=(-10/7)*a,x3=a}`;**

`> `

or

`> `
**S = span(matrix(3,1,[-5/7,-10/7,1]));**

or

`> `
**S = span(matrix(3,1,[-5,-10,7]));**

`> `

Is S a subspace of ?

Geometrically, S represents a line through the origin. The set S is closed under addition and

scalar multiplication. Therefore, S is a subspace. Hence, the solution set of a homogeneous

system is a subspace. This is called the
__nullspace__
of the matrix A and is denoted by N(A).

Is the solution set of a nonhomogeneous system of linear equations a subspace?

*****************************************************************

The
__ROW SPACE__
of an nxm matrix A is the space spanned by the rows of the matrix A.

Since the span of a set of vectors is a subspace, the row space of an
*nxm*
matrix A is a

subspace of .

The
__COLUMN SPACE__
of an nxm matrix A is the space spanned by the columns of the

matrix A.

Since the span of a set of vectors is a subspace, the column space of an
*nxm*
matrix A is a

The
__NULL SPACE__
of an nxm matrix A is the solution set to Ax=0.

*****************************************************************

**Example 10: **
Find the row space, column space, and the null space.

`> `
**A:=matrix([[1,8,3],[6,-3,7]]);**

`> `

`> `
**AUG := matrix([[1,8,3,0],[6,-3,7,0]]);**

`> `
**solve({x1+8*x2+3*x3=0,6*x1-3*x2+7*x3=0},{x1,x2,x3});**

`> `

************************************************************

A linear system Ax=b has a solution if and only if b is in the column space of A.

************************************************************

**Bases**

*********************************************************************

Let W be a subpsace of
. A subset {
,. . .,
} of W is a
** basis**
for W if

every vector in W can be expressed
* uniquely*
as a linear combination of
,. . .,
.

***********************************************************************

**Example 11:**
Consider the two independent vectors

`> `
**e1:=vector([1,0]); e2:=vector([0,1]);**

`> `

Is every vector in a linear combination of the vectors and ?

Let w be any vector in

`> `
**w:= vector([x,y]); **

`> `

If w = [x,y] is a linear combination of and

`> `
**evalm(w)=c1*evalm(e1)+c2*evalm(e2);**

Then = x and = y. Hence w is

`> `
**evalm(w)=x*evalm(e1)+y*evalm(e2);**

`> `

Therefore every vector in can be written uniquely as a linear combination of the two

vectors and . Thus the set { , } is spans and is a basis for .

**Example 12: **
Now
** **
consider the vectors

`> `

`> `
**e1:=vector([1,0]);e2:=vector([0,1]);
u:=vector([2,5]);**

`> `

Is the set { , , u} a basis for ? Let w be any vector in

`> `
**w:=vector([x,y]);**

`> `

If w = [x , y] is a linear combination of , and u

`> `
**evalm(w)=c1*evalm(e1)+c2*evalm(e2)+c3*evalm(u);**

`> `

This yields the system of equations

`> `
**eq1:=c1+2*c3=x; eq2:=c2+5*c3=y;**

`> `

On solving the system we get

`> `
**solve({eq1,eq2},{c1,c2,c3});**

`> `

Since the system is consistent, the vectors , , and u do span . However,

the linear combination is not unique.

`> `
**evalm(w)=(x-2)*evalm(e1)+(y-5)*evalm(e2)+evalm(u);**

**************************************************************

The set {
,. . . ,
} is a
** basis**
for W = sp{
,. . . ,
} in

if and only if the zero vector is a
*unique*
linear combination of the
that is

if and only if + . . . + = 0 implies that

= = . . . = = 0.

***************************************************************

We call the set {
, . . . ,
} the
** standard basis**
for
.

**Example 13:**
Consider the vectors

`> `

`> `
**v1:=vector([0,1,1]); v2:=vector([1,1,2]);v3:=vector([1,0,2]);**

`> `
**eq1:=c2 + c3=0: eq2:=c1+ c2=0: eq3:= c1+2*c2+2*c3=0:**

`> `
**print(eq1); print(eq2); print(eq3);**

`> `
**solve({eq1,eq2,eq3},{c1,c2,c3});**

`> `

Change vector v1

`> `
**v1 := vector([3,-2,6]);**

`> `
**eq1:= 3*c1 + c2 + c3=0: eq2:=-2*c1+ c2=0: eq3:= 6*c1+2*c2+2*c3=0:**

`> `
**print(eq1); print(eq2); print(eq3);**

`> `
**solve({eq1,eq2,eq3},{c1,c2,c3});**

`> `

**Unique Solutions for Ax=b**

*****************************************

Let A be a square matrix, i.e.
*A*
is a
*n x n*
matrix. The following are equivalent:

1.) The linear system
*Ax=b*
has a unique solution

2.) The matrix
*A*
is row equivalent to the identity matrix I.

3.) The matrix
*A*
is invertible.

4.) The column vectors of
*A*
form a basis for

*******************************************

*******************************************

Let
*A*
be an
*m x n*
matrix. The following are equivalent.

1.) Each consistent system
*Ax=b*
has a unique solution

2.) The reduced row-echelon form of
*A*
consists of the
*n x n*
identity matrix

followed by m-n rows of zeros

3.) The columns of
*A*
form a basis for the column space
*A*
.

*******************************************

**Exercises**

1, 3, 9, 12, 14, 17, 19, 23, 25, 29, 30,31, 38.